C/P UWorld Q's

अब Quizwiz के साथ अपने होमवर्क और परीक्षाओं को एस करें!

Steroisomers

Multiple peaks with the same m/z in the mass spectrum of an unsaturated molecule could correspond to cis/trans isomers of the molecule; Docosenoic acid is an unsaturated fatty acid, meaning it contains a disubstituted double bond, and can be in the cis or trans form. Both isomers have the same molecular weight and m/z ratio, but will come off the GC column at different times, yielding two distinct peaks. Therefore, it can be inferred that Peaks A and B are cis/trans isomers of docosenoic acid.

Velocity

Negative sign for velocity indicated the object is traveling backwards, not that it is slowing down.

In which compound does manganese have a +6 oxidation number? A) MnO B) MnO2 C) K2MnO4 D) KMnO4

C- Using the oxidation state rules for the formula K2MnO4 shows that manganese has an oxidation state of +6. Although manganese is a variable metal that has seven possible oxidation states, the states of oxygen (−2) and potassium (+1) are consistent and can be used to find the value for manganese. Four oxygen atoms (−2 each) and 2 potassium atoms (+1 each) sum to −6. For the net charge of the formula to be zero, the manganese atom must have a value of +6.

Which of the following atoms will be expected to have the smallest second ionization energy? A) Na B) C C) O D) Ca

D - Metals have lower ionization energies than non-metals as long as the ionization event involves a valence electron. Since Na is an alkali metal, it has only one valence electron and has a large second ionization energy. Ca is an alkaline earth metal and has two valence electrons. It will therefore have the smallest second ionization energy of the four atoms listed, which include Na and two non-metals.

The fundamental, resonant wavelength of a pipe open at both ends that is 1 m long and 0.1 m in diameter is: A) 0.1 m. B) 0.2 m. C) 1.0 m. D) 2.0 m.

D - Pipes and tubes have their resonant wavelengths when standing waves develop. An open pipe has its fundamental, resonant wavelength at twice the length of the pipe. Both ends have displacement antinodes (maximum amplitudes) with a node in the middle of the pipe.

Which of the following best describes the bonds between Cu2+ and the nitrogen atoms of the ammonia molecules in [Cu(NH3)4]2+? A) Ionic B) Covalent C) Coordinate ionic D) Coordinate covalent

D - The Lewis acid-base interaction between a metal cation and an electron pair donor is known as a coordinate covalent bond.

After a block began to slide, how did its speed vary with time? (Note: Assume that the tension and kinetic friction forces on the block were constant in magnitude.) A) It was constant in time. B) It increased exponentially with time. C) It was first constant, then increased linearly with time. D) It increased linearly with time.

D - The coefficient of kinetic friction is always lower than that of static friction. Therefore there is a net accelerating force on the block once it starts to slide. A constant force on a mass produces a constant acceleration (Newton's second law). Thus, the velocity of the block increases linearly with time.

Which of the following reaction conditions would encourage a reaction to proceed in the forward direction? A) A reaction that has a reaction quotient lower than its equilibrium constant B) A reaction that has a reaction quotient higher than its equilibrium constant C) A reaction that has a reaction quotient equal to its equilibrium constant D) A reaction that has a reaction quotient equal to zero

A

Amide VS Enamine

Amides are carboxylic acid (CA) derivatives made up of a carbonyl bonded to a nitrogen atom. These compounds form from the condensation of any CA derivative with a primary or secondary amine; while an enamine is a functional group that contains nitrogen linked to a carbon that is double-bonded to another carbon and is not a CA derivative. It is formed by the addition of a secondary amine to an aldehyde or ketone.

What feature of dansylalanine allows for monitoring protein unfolding? A) Isomerization of double bonds B) Conjugated pi bonds C) Resonance structure at the C-terminus D) Chirality of the amino acid

B - According to the passage, noncanonical amino acids such as dansylalanine can be used to monitor protein structure because they are fluorescent. Fluorescence occurs when a photon in the ultraviolet (UV) or visible region is absorbed by a fluorophore; For a molecule to absorb UV or visible light, its electrons must be sufficiently delocalized so that the energy difference between the ground and excited states corresponds to the energy of the light being absorbed. This delocalization can be achieved by a highly conjugated system, or a system with several alternating single and double bonds. In general, a larger system corresponds to a higher wavelength absorbed. The π electrons can be delocalized through the p-orbitals, creating an extended network of conjugated pi bonds known as resonance stabilization.

Which of the following best describes the movement of an electron after it is ejected from the cathode? A) It is stationary until collisions propel it toward the anode. B) It moves with constant speed toward the anode. C) It accelerates toward the anode. D) It exits through a side of the vacuum photodiode.

C - A charged particle accelerates in an electric field. The electron starts with a velocity that increases as it approaches the anode through the vacuum

Platinum most likely increases the reaction rate by stabilizing: A) The reactants B) The products C) The intermediate step D) The transition state.

D - A catalyst is a substance that increases the rate of a reaction without being consumed by the reaction. It increases the rate by stabilizing the transition state and thereby lowering the activation energy of the reaction. The passage states that platinum increases the reaction rate of hydrogen combustion without being consumed; therefore, platinum must be acting as a catalyst, stabilizing the transition state of the combustion reaction.

Electron Affinity

Electron affinity trends show that apart from some exceptions caused by atomic size and orbital-filling effects, values tend to become more negative (higher affinity) moving left to right across a period and become more positive (lower affinity) moving down a group on the periodic table. The data in Figure 2 indicate that nonmetals tend to have the most negative values for electron affinity. Therefore, except for nitrogen and the noble gases (which require energy), adding an electron to a nonmetal is easier than adding an electron to a metal.

Isoelectronic

If two atoms are isoelectronic, they have the same number of electrons.

Le Chatelier's Principle

Increasing the pressure of a system will shift the reaction to the side of fewer moles of gas, whereas decreasing the pressure will shift it to the side of greater moles of gas.

Maltose

Maltose is composed of two glucose molecules that are connected by an alpha-1,4 linkage.

While carbon dioxide exhibits 4 vibrational modes in space, its experimentally obtained IR spectrum shows no more than two distinct peaks. This is partly because: A) The two oxygen nuclear exhibit nuclear spins that are equal in magnitude but opposite in direction. B) Three vibrational modes are considered energetically degenerate. C) One of the vibrational mode is not observable by IR spectroscopy. D) Two of the vibrational modes are not excitable by IR spectrometers of conventional power.

The symmetric stretching mode of this molecule does not induce a dipole. It is that's effectively invisible to ir spectroscopy

In a nearsighted individual, the image of a distant object is focused: A) In front of the retina, requiring diverging lens correction. B) In front of the retina, requiring converging lens correction. C) Behind the retina, requiring diverging lens correction. D) Behind the retina, requiring converging lens correction.

A - A nearsighted (myopic) person can focus on nearby objects but cannot clearly see objects far away. The nearsighted eye has a focal length that is shorter than it should be, so the rays from a distant object form a sharp image in front of the retina. This condition is corrected by glasses with a diverging lens.

A 7-N force and an 11-N force act on an object at the same time. Which of the following CANNOT be the magnitude of the sum of these forces? A) 2 N B) 8 N C) 12 N D) 18 N

A - The largest net force occurs when the forces point in the same direction; then, the magnitudes will add mathematically to a value of 18 N. Similarly, the smallest net force occurs when the forces point in opposite directions and the lesser magnitude is subtracted from the greater; this results in a value of 4 N. Of the answers listed, only A (2 N) does not fall between these values. Thus, A is the best answer.

The next generation of lithium ion batteries, so-called lithium air batteries, utilize lithium metal for the anode and an inert cathode exposed to air. Which of these common gases present in the air is likely to act as the electrochemically active portion of the cathode? A) O2 B) N2 C) Ar D) H2

A -02 readily undergoes reduction in reactions such as combustion, while N2 is considered largely inert.

Phase Diagram Interpretation

A phase diagram shows a substance's stability in each phase as a function of pressure (y-axis) and temperature (x-axis). The boundary lines between two phases indicate the conditions under which they are in equilibrium. According to the passage, Raoult law states that the addition of any solute to a pure liquid lowers the freezing point and vapor pressure of the resulting solution. A substance's phase diagram shows this change by moving the phase boundaries to lower temperatures and pressures, respectively.

Bronsted Lowry Acid/Base

According to the Brønsted-Lowry definitions, an acid is a chemical species that donates a proton (H+ ion), and a base is a chemical species that accepts a proton. The conjugate of an acid (or base) is the species that the acid (or base) becomes after losing (or gaining) one proton. A base is converted into its conjugate acid when the base accepts a proton. Likewise, an acid is converted into its conjugate base when the acid donates a proton. Therefore, an acid (or base) and its conjugate have chemical formulas that differ by only one proton. Whether a molecule functions as an acid or as a base sometimes depends on the relative acidity (acid strength) of the molecule. For example, H2SO4 and HNO3 both usually function as strong acids (proton donors). However, if H2SO4 and HNO3 are mixed together, H2SO4 is demonstrated to be a stronger acid than HNO3 because the HNO3 is induced to acceptanother proton to form H2NO3+. By the Brønsted-Lowry definition, a base is a proton acceptor. Therefore, in the forward reaction of the equilibrium, HNO3 acts as a Brønsted-Lowry base by accepting a proton from H2SO4.

Shiny nickel metal granules added to orange liquid bromine mixed in alcohol produced a combination reaction yielding a blue-green solution of nickel(II) bromide. In the reaction between Ni and Br2: A) Bromine forms a chelate with nickel B) Nickel undergoes an oxidation-reduction reaction with bromine C) The reaction forms a precipitate of nickel(II) bromide D) Nickel and bromine participate in a Brønsted-Lowry acid-base neutralization

B

The properties of sound waves allow them to propagate through: A) gases only. B) liquids and gases only. C) solids, liquids, and gases only. D) vacuum, solids, liquids, and gases.

C - The question requires you to identify a major property of sound waves. For sound waves to propagate, there must be a physical medium consisting of atoms that can interact with each other and therefore propagate the perturbation produced by the sound wave. Such medium can be in either solid, liquid, or gaseous state. Recalling one example of each medium is sufficient to determine the correct answer.

In which of the following does sound travel most rapidly? A) Air (0°C) B) Water (10°C) C) Iron (20°C) D) Sound travels at approximately the same speed in all of the above.

C - The speed of sound is greatest by far in a solid because of the strong intermolecular bonds and close proximity of the molecules. Because the temperatures are close to each other (and even favor iron slightly), iron will have the highest sound speed. Thus, C is the best answer.

In Reaction 1 shown in the passage, the oxidizing agent is: 2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O A) K2MnO4, because potassium loses electrons. B) MnO2, because manganese gains electrons. C) KOH, because hydrogen loses electrons. D) O2, because oxygen gains electrons.

D

When comparing single, double, and triple covalent bonds formed between two atoms X and Y, decreasing the number of π bonds between the atoms will: I. Decrease the overall bond dissociation energy and bond strength. II. Increase the bond length. III. Decrease the bond rigidity (allow more bond rotation). A) I and II only B) I and III only C) II and III only D) I, II, and III

D

Which statements correctly describe electrolysis? During electrolysis: I. An electrochemical decomposition reaction occurs II. Electric current is supplied as the energy to perform an endergonic reaction III. The electrochemical cell operates with a negative ΔG IV. Electric current flows in a nonspontaneous direction within an electrochemical cell. A) I and IV only B) I, II, and III only C) II and IV only D) I, II, and IV only

D

Vectors

Displacement and velocity are vectors, meaning they have both direction and magnitude. In contrast distance, speed, and work or scalars, they only convey magnitude with no indication of the direction in which the work was performed.

Lactose

Glucose and galactose are the monosaccharide components of the disaccharide lactose. They are connected by a beta-1,4 acetal linkage and is cleaved by lactase.

Mass Spec

Mass spec requires mobilization of the sample in the gas phase, which is typically accomplished via bombardment with ions. While the actual degree of damage to standby a protein sample during this process is up for debate it constitutes a reasonable concern. Additionally, molecules pass through a mass back in a vacuum and are stripped of their surrounding water molecules or like lipid molecules. Although no consequences have been yet reached it is reasonable to suggest that such a dramatic deviation from cellular conditions could alter the structure of a protein under observation

Glucose Anomers

The beta anomer of glucose is slightly more stable than the alpha anomer. When provided with a catalyst, such as an acid, the alpha anomer will tend to mutarotate into the beta anomerr until an approximate 1:2 anomer ratio is achieved.

Lenses

The lenses have a negative focal length which means they are diverging lenses. Such lenses form virtual and reduced images of objects situated at distances larger than the focal length.

Visible light travels more slowly through an optically dense medium than through a vacuum. A possible explanation for this could be that the light: A) Is absorbed and re-emitted by the atomic structure of the optically dense medium. B) Is absorbed and re-emitted by the nucleus of the material in the optically dense medium. C) Bounces around randomly inside of the optically dense medium before emerging. D) Loses amplitude as it passes through the optically dense medium

A - A is known to occur and is the reason for the slowing down of light. B is incorrect because the nucleus is involved. C is incorrect because the motion of the photons is certainly not random. D is true but does not answer the question.

All the following statements regarding the sapontification of triacylglycerols are true EXCEPT: A) It yields two fatty acid salts and a glycerol molecule. B) It is the reverse of the esterification reaction between fatty acids and glycerol. C) It is a hydrolysis reaction. D) Utilizes OH- ions for nucleophilic attack on carbonyl carbon.

A - A triacylglycerol consists of a glycerol backbone bound via ester linkages to three fatty chains. Thus, the saponification of such a molecule will yield three, not two, fatty acids cells and one glycerol molecule.

Which single bond present in nitroglycerin is most likely the shortest? A) C-H B) C-O C) C-C D) O-N

A - All of the bonds listed are single bonds. Since hydrogen has a much smaller atomic radius than second period elements, the covalent bond between C and H is shorter than any of the other bonds listed

Considering the type of element represented by each of the following, which of the elements below is most likely to have the highest electrical conductivity? A) Ca B) Si C) Se D) I

A - Ca is a metal, while Si is a metalloid, and Se and I are nonmetals; look at periodic table

Which of the following accurately describes the thermodynamic products of a reaction? I. They have a lower free energy than the kinetic products II. They have a lower activation energy than the kinetic products III. They are more stable than the kinetic products IV. They tend to form at higher temperatures than the kinetic products A) I and II B) I, III, and IV C) II, III, and IV D) II and III

B

When switch S is closed to the left, charge begins to accumulate on the capacitor. Charge cannot accumulate indefinitely because: A) The variable resistor inhibits the current flow. B) The battery continually loses charge. C) Successive charges brought to the plates are repelled by charges accumulated earlier. D) The fixed resistor loses energy to heat.

C - The two plates of the capacitor collect charges of opposite sign. As more charge arrives it is harder and harder to fill the plates until finally an equilibrium occurs, thus C is correct.

Disproportion Reactions

One element can undergo both oxidation and reduction. In this case, atoms from the same element simultaneously act as the oxidizing agent and the reducing agent.

Disproportionation Reactions

In a disproportionation reaction, the same element (at a given oxidation state) undergoes both oxidation and reduction, with some of the atoms being oxidized and other atoms of the same element being reduced. Comparing the oxidation number for each element in the reactants with the oxidation number for the same element in the products enables the identification of a disproportionation.

By what factor is the hydronium ion concentration increased in the stomach lumen compared to the parietal cell? A) 10^−6 B) 6 C) 106 D) 10^−1

Moving from the parietal cell (pH 7.0) to the stomach lumen (pH 1.0) involves a change of 6 pH units. The increase in [H3O+] going from the parietal cell to the stomach lumen can be determined as follows: Factor Δ[H3O+] = 10−(1 − 7) = 10^6 Therefore, the [H3O+] in the stomach lumen is one million (106) times stronger than the [H3O+] in the cell cytosol.

Bond Energies

Sigma bonds are lower in energy and more stable than pi bonds, and therefore have a higher dissociation energy. Sigma bonds form first between two atoms, and every bond formed thereafter is a pi bond. Although individual pi bonds are weaker than sigma bonds, a double bond is composed of both a sigma and a pi bond, and therefore is stronger than a single bond.

Because step 2 of the reaction in aqueous phase is rate-limiting, the formation of ROI from tyrosine and ozone in step 1 can be assumed to reach a state of equilibrium. Tyrosine(aq)+ O3(aq)⇌ ROI(aq) + HO3(aq)Tyrosine(aq)+ O3(aq)⇌ ROI(aq) + HO3(aq) Which of the following will occur if the temperature of this system is raised from 298.15 K to 350 K? A) Keq will increase B) The reaction will be driven toward the reactants C) The reaction will be driven toward the products D) Keq will remain the same.

A - Equilibrium is reached in a reaction when the rate at which products are formed is equal to the rate at which reactants are formed. Per Le Châtelier principle, if a system in equilibrium is disturbed by a change in temperature, pressure, or a component's concentration, the system will counteract the disturbance by driving the reaction in a given direction such that equilibrium is restored. Changes in pressure and concentration achieve this without altering the equilibrium constant Keq, whereas a change in temperature leads to a change in Keq. The effect of temperature on an equilibrium reaction can be illustrated by treating heat as a chemical reagent and applying Le Châtelier principle. Step 1 of the nitration of tyrosine in an aqueous environment is an exothermic reaction, and heat can be viewed as a product of the reaction. Raising the temperature causes a shift toward the reactants as heat is consumed to rebalance the equilibrium reaction. The shift toward reactants leads to an increase in reactant concentration, and therefore a decrease in Keq.

Which of the following molecules is unable to form hydrogen bonds with water? A) C2H4Br2 B) C3H9N C) C9H18O2 D) C2F4

A - Hydrogen bonding is a strong intermolecular force that occurs between a hydrogen atom on one molecule and an electronegative atom with a lone electron pair on another molecule. The hydrogen atom is the bond donor and must be covalently attached to a small electronegative atom that withdraws electrons from hydrogen, giving it a partial positive charge. The partial positive charge is then attracted to the partial negative charge of the lone pair of electrons in the bond acceptor. These interactions can only occur over short distances, so they require sufficiently small electronegative atomsin both the donor and acceptor molecules, and are essentially limited to fluorine, nitrogen, and oxygen. Carbon is a small atom but not sufficiently electronegative to participate in hydrogen bonding. On the other hand, bromine is electronegative but too large to participate in hydrogen bonding. Therefore, C2H4Br2 is unable to form hydrogen bonds with water.

Moving down the column, the alkaline-earth metals are observed to give increasingly vigorous reactions when forming ionic bonds with nonmetals (Reactions 1-3). Based on atomic properties, this trend in reactivity is best explained by comparing: A) The energy required to remove an electron from each atom. B) The tendency of each atom to attract electrons within a bond. C) The extent to which the electron cloud of each atom can be distorted by an external charge. D)The energy released when an electron is added to each atom.

A - Ionization energy is the energy required to remove an electron from an atom. Because of the associated ionization and electron transfer involved in forming ionic bonds, the reactivity of atoms forming ionic compounds increases as the ionization energy decreases.

A student has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact melting temperature. If the student lights a Bunsen burner and holds it for a fraction of a second under the beaker, what will happen to the metal? A) A small amount of the metal will turn to liquid, with the temperature remaining the same. B) All the metal will turn to liquid, with the temperature remaining the same. C) The temperature of the metal at the top of the beaker will increase. D) The temperature of the whole mass of the metal will increase slightly.

A - Melting occurs at a constant temperature because a certain amount of energy, the latent heat of fusion, is needed to convert a substance from its solid to liquid state. The temperature of the metal will not increase above its melting point until all of the metal has melted. The small amount of heat supplied by the bunsen burner is insufficient to melt 100 g of the metal but could melt a small amount of the metal at the constant temperature of the melting point. Thus, answer choice A is the best answer.

Suppose that Experiment 1 was done in diethyl ether instead of acetone, and NaI was replaced with another nucleophile. The conjugate base of which of the following compounds would be the most nucleophilic? A)CH3(CH2)2CH3, pKa = 50 B) (CH3)2NH, pKa = 40 C) CH3CH2OH, pKa = 15.9 D) HF, pKa = 3.2

A - Nucleophiles consisting of a negatively charged nitrogen, oxygen, or fluorine contain charged atoms that are more electronegative and have a stronger hold on their electrons, making these species less nucleophilic than carbon nucleophiles. When comparing the nucleophilicity of atoms of equal negative charge, nucleophilicity tends to increase from right to left across a row of the periodic table as electronegativity decreases. Atoms of higher electronegativity more effectively stabilize negative charge and less readily donate electron density to electrophiles whereas those of lower electronegativity stabilize a negative charge less effectively and more readily donate electrons to electrophiles.

If three equivalents of NaOH are used for triacylglycerol saponification, the triacylglycerol will be: A) Completely hydrolyzed, because one equivalent of NaOH is needed to hydrolyze each ester. B) Completely hydrolyzed, because a catalytic amount of NaOH is needed to hydrolyze the entire triacylglycerol. C) Partially hydrolyzed, because two equivalents of NaOH are needed to hydrolyze each ester. D) Partially hydrolyzed, because six equivalents of NaOH are needed to hydrolyze the entire triacylglycerol.

A - Saponification is the hydrolysis of an ester with a strong base. Triacylglycerols contain three fatty acids bonded to a molecule of glycerol through ester linkages, and saponification of a triacylglycerol with a strong base releases free fatty acids (as sodium salts) and a molecule of glycerol. One equivalent of base is needed to hydrolyze one ester linkage; therefore, three equivalents of base are needed to completely hydrolyze a triglyceride.

A 10 KG boxes sit on a cargo round with a 60° incline. In order for the box remain in place, which of the following relationships must be true? A) The coefficient of static friction for the ramp must be at least 1.7. B) The coefficient of kinetic friction for the ramp must be at least 1.7. C) The coefficient of static friction cannot exceed a value of one, so the ramp must be lowered to the smaller angle in order for the box to be equilibrium. D) The coefficient of kinetic friction cannot exceed a value one, the rate must be lower to the smaller angle in order for the box to be at equilibrium.

A - Static friction equation to find value. B, D: kinetic friction can only be applied on the object of sliding against surface in question

The C=S double bond in carbon disulfide (CS2) consists of a combination of one σ bond and one π bond. Compared to the bond dissociation energy (bond strength) of the π bond, the bond dissociation energy of the σ bond portion of the C=S bond is: A) Greater B) Less C) The same D) Proportional

A - Stronger bonds require more energy to break (dissociate) than weaker bonds. The overall strength of a bond results from the sum of all σ and π bonding contributors. As a result, when comparing bonds involving the same two types of atoms, a triple bond is stronger than a double bond, and a double bond is stronger than a single bond because double and triple bonds are composed of both σ and π bonds. However, if the strengths of the σ bond and π bond contributors are considered separately, π bonds are weaker than σ bonds. The end-to-end orbital overlap in σ bonds is more efficient than the side-to-side orbital overlap in π bonds. This causes σ bonds to exist in a more stable, lower energy state. As a result, breaking a σ bond requires more added energy than breaking a π bond (ie, a σ bond has a greater dissociation energy).

If researchers used the Gabriel synthesis to make the dansylalanine backbone, which of the following statements is true? A) Potassium phthalimide is a starting material. B) Diethyl malonate is a starting material. C) Potassium cyanide is a starting material. D) An aldehyde is a starting material.

A - The Gabriel synthesis, also known as the malonic ester synthesis, is a method used to make primary amines, including α-amino acids, without overalkylation of the amine. The amine is generated from potassium phthalimide, a "protected" form of ammonia that prevents multiple alkylations due to the steric hindrance of the phthalimide group. The synthesis begins with an SN2 reaction, where potassium phthalimide is the nucleophile that attacks an alkyl halide. The amine is then "deprotected" by removal of the phthalimide group.

When the number of photons incident on the cathode with energies above the value of the work function increases, which of the following quantities also increases? A) Number of electrons ejected B) Potential energy of each ejected electron C) Magnitude of the electric field between the electrodes D) Speed of electrons at the anode

A - The number of incident photons affects only the number of electrons, not their energies. The electron energies depend on photon energy, the cathode work function, and the potential difference between the cathode and anode.

In an aqueous solution of 1 × 10−3 M HCN (Ka ≈ 1 × 10−9), HCN can best be described as a: A) Weak acid, because [H+] < 1 × 10−3 M. B) Strong acid, because [H+] = 1 × 10−6 M. C) Weak base, because [H+] < 1 × 10−7 M. D) Strong base, because [H+] ≈ 1 × 10−3 M.

A - The small Ka value for HCN indicates that the non-ionized form is favored, and it is a weak acid. Evaluation of the [H+] gives [H+] = 1 × 10−6 M. This [H+] is much less than the HCN concentration (1 × 10−3 M) and confirms that only a small percentage of the dissolved HCN molecules in solution are ionized (consistent with a weak acid). Although the nitrogen atom of the cyanide group does have an available lone pair of electrons that could act as a weak proton acceptor (weak base), HCN is donating protons (acting as a weak acid). The calculated value of [H+] = 1 × 10−6 M (pH 6) is greater (more acidic) than a neutral solution ([H+] = 1 × 10−7 M, pH 7) and confirms that HCN acts overall as a weak acid rather than as a weak base.

The glycerol backbone of a triacylglycerol is attached to three fatty acids via: A) Ester bonds. B) Disulfide linkages. C) Carbon=carbon double bonds. D) Ether linkages.

A - To form a triacylglycerol molecule from its component parts, three free fatty acids must be joined with glycerol via an esterification reaction. Technically, triacylglycerol is a triester.

Compared to IR spectroscopy, ultraviolet visible spectroscopy involves: A) Induction of electronic excitations, and consequently exposes the sample to higher energy light. B) Involves the induction of electronic relaxations, and consequently exposes the samples to lower energy light. C) Relies on the unique vibrational energies a specific chemical bonds, requiring a higher incidence energy. D) Relies on the unique vibrational energies of specific chemical bonds, requiring a lower incidence energy

A - UV-Vis spec involves exposure of a sample to incident light sufficient to promote electrons of the highest occupied molecular orbital to the lowest and occupied molecular orbital. As its name implies, the incident light lies in the UV visible region, which includes waves of higher energy, and shorter wavelength, then the infrared region.

Which of the following statements best explains why hemoglobin is red when it binds to O2? A) O2 interacts with iron's d orbitals B) O2 is a nonpolar ligand C) O2 has lone pairs of electrons D) O2 changes the surrounding protein structure.

A - When placed within a coordination sphere, the atomic d orbitals of a metal are no longer degenerate, so they will have different energies. This occurs because some orbitals are pointing toward the ligands, which are electron donors, and other d orbitals are pointing away from or between the ligands. The orbitals that point toward the ligands are higher in energy than the orbitals pointing away from or between them because electrons repel each other. The energy of the orbitals determines which wavelengths of light can be absorbed. Some ligands cause greater differences in the energy of the d orbitals than other ligands, resulting in the absorption of different wavelengths of light. These wavelengths are usually in the visible spectrum. Oxygen's lone pairs are a factor in the difference in energy in iron's d orbitals, but the red color is due to electron transitions between the d orbitals, not the lone pair of electrons on O2.

What is the expected result of increasing the pressure the given reaction occurs at? N(g) + O2(g) ⇋ N2O5 (g) A) The reaction will shift right B) The reaction will shift left C) There will be an increase in the concentration of elemental oxygen D) Increasing the pressure will not affect the reaction

A - When the pressure is increased, a gaseous reaction shifts to the side of fewer moles.

Sound of a known frequency, wavelength, intensity, and speed travels through air and bounces off an imperfect reflector which is moving toward the source. Which of the following properties of the sound remains the same before and after reflection? A) Speed B) Intensity C) Frequency D) Wavelength

A - Within still air, the speed of sound remains constant. Thus, A is the best answer.

If energy density is the amount of energy per unit mass, which of the following statements must be true? A) NiCd batteries have a higher energy density than lead storage batteries B) Lead storage batteries produce less energy than NiCd batteries C) NiCd batteries produce more electrons per mole of metal than lead storage batteries D) Both lead storage and NiCd batteries have the same energy density

A - work that is done per unit of charge, or joules of work per coulomb of charge. The amount of work that a battery does is due to the conversion of chemical energy into electrical energy (joules) by an oxidation-reduction reaction. Energy density relates the E°cell of a battery to its mass, in which a battery with a high energy density has a high energy output, a low weight, or both. According to the passage, one of the advantages of NiCd batteries is that they have a higher energy-to-weight ratio than lead storage batteries. Even though lead-acid batteries have a higher E°cell, NiCd batteries have more energy per unit mass and are smaller in size. Therefore NiCd batteries have a higher energy density. (Choice B) Lead storage batteries provide about 2.0 V of charge whereas NiCd batteries provide 1.3 V. The amount of energy a cell provides is directly proportional to the potential E°cell, so lead storage batteries produce more energy in absolute terms than NiCd batteries. (Choice C) Based on the oxidation-reduction half-reactions for the NiCd battery and the lead storage battery, both reactions produce 2 moles of electrons per 1 mole of Cd or Pb that is oxidized. (Choice D) Lead storage batteries have a lower energy-to-weight ratio compared to NiCd batteries, meaning lead storage batteries are heavier relative to the amount of energy they produce. Therefore, they do not have the same energy density.

Tollen's Test

A positive talons test is indicated by the formation of a silver mirror, or shiny AG precipitate. This requires the presence of a reducing sugar.

An electrolytic cell is designed to produce pure copper from CuSO4. An increase in which of the following cell conditions will most effectively increase the rate at which pure copper is produced? A) The concentration of SO42+(aq) B) The current of electricity C) The size of the cathode D) The size of the anode

B - An electrolytic cell drives a non-spontaneous reaction using electrical energy. In the electrolytic cell, electrons are driven in through the cathode and drawn out through the anode. The rate at which this occurs is the current. By increasing the current, electrons are forced in faster, increasing the rate of reduction, and are withdrawn faster thus increasing the rate of oxidation.

Cellulose differs from starch in that cellulose includes an: A) Alpha-1,4 acetal linkage. B) Beta-1,4 acetal linkage. C) Alpha-1,6 acetal linkage. D) Beta-1,6 acetal linkage.

B - Cellulose contains beta-1,4 acetal linkages between glucose molecules. Humans lack an enzyme to clear these bonds, which explains why we cannot digest cellulose. Starch contains alpha-1,4 acetal linkages, which we can break down

Ethyl acetate could be better separated from ethanol by doing which of the following? A) Replacing helium with nitrogen as the carrier gas B) Running the mixture through a longer column C) Starting the chromatograph at a higher temperature D) Increasing the carrier gas flow rate

B - Chromatography methods separate molecules based on their relative affinities for a stationary phase versus a mobile phase. For compounds to separate efficiently on any column, they need sufficient time to interact with the stationary phase. Given enough time, even subtle differences in affinity for the stationary phase can be amplified, allowing separation of compounds with similar properties. Increasing the length of the stationary phase through which the compounds must travel can provide the necessary time. Ethanol and ethyl acetate have similar, but not identical, boiling points and can be separated on a gas chromatograph by running them through a longer column.

The effective nuclear charge experienced by the valence electrons of a neutral nitrogen atom is: A) +7 B) +5 C) −3 D) −5

B - Each proton in the nucleus of an atom contributes one unit of positive charge that exerts an electrostatic attraction on the negatively charged electrons around the atom. However, in atoms with several electrons, the core electrons positioned between the valence electrons and the nucleus provide a shielding constant S that counteracts part of the full nuclear charge Z attracting the valence electrons. As a result, the valence electrons experience an effective nuclear charge Zeffthat is less than the full nuclear charge, such that Zeff=Z−SZeff=Z-S Precisely calculating S is difficult and depends on the electron density of a given electron orbital configuration, but a rough, first-order approximation can be made by assuming that S is approximately equal to the number of core electrons. When comparing atoms within the same period (row) of the periodic table, this approximation shows that Zeff increases as the atomic number increases. However, because this approximation does not account for electron density or orbital distance, it is less useful for precisely comparing Zeff between atoms in the same group (column) or in different rows of the periodic table. Applying this approximation to calculate the Zeff experienced by the valence electrons of a nitrogen atom with 7 protons (Z = 7) and 2 core electrons (S = 2) gives: Zeff,N=Z−S=(7−2)=+5Zeff,N=Z-S=7-2=+5

Oxygen is the most electronegative element of Group 6A. However, it also has the lowest electron affinity in the group. What factor can be used to explain this discrepancy? A) Oxygen is less stable than the other Group 6A elements B) Oxygen's valence electrons experience more electron-electron repulsion than other Group 6A elements C) Oxygen's valence electrons are more shielded from the positively charged nucleus than other Group 6A elements D) Oxygen does not have d orbitals available for additional electrons

B - Electronegativity is a qualitative measure of how strongly electrons are attracted to an atom within a bond. Group 6A (Group 16) elements follow the trend that electronegativity decreases going down the group on the periodic table. When bonded to another atom, oxygen strongly attracts electrons toward itself, creating unequal electron sharing that produces a dipole moment. In contrast, electron affinity (EA) is the quantitative measure of the energy change when an electron is added to an atom in the gas state. The more negative the change in energy, the more energy released upon addition of an electron. Although EA generally increases (becomes more negative) from left to right across a period, EA generally decreases(becomes more positive) going down a group because the electron is added at an increased distance from the nucleus. The exception to this trend is when an atom is very small (ie, first or second period). Small atoms such as oxygen and fluorine have several electrons crowded around a small nucleus, resulting in greater electron-electron repulsion. Adding another electron would increase repulsion forces even further, so it is more difficult to add an electron to oxygen than to a larger atom with lower repulsion forces.

Unlike IRS spectroscopy, UV-Vis spectroscopy is: A) Reliant on the use of a longer wavelength light. B) Reliant on the use of higher energy radiation. C) Less likely to be used to detect molecules like dodeca- 2,4,6,8- tetradiene. D) More likely to separate compounds on the sole basis of net charge

B - IR spec utilizes infrared radiation, while UV viz. uses ultraviolet light. UV radiation is a higher frequency, shorter wavelength, and a higher energy than infrared rays

When comparing atoms across the same row of the periodic table, which of the following groups will contain the atom with the lowest second ionization energy? A) Group 1 B) Group 2 C) Group 14 D) Group 16

B - If comparing elements in period (row) three, the nonmetals Si (Group 14) and S (Group 16) are farther to the right in the period, and both have a higher second ionization energy than Mg. However, the second ionization energy of Na (Group 1) is much higher than that of Mg (Group 2) because Na has only one valence electron and removing a second electron from Na requires the loss of a core electron. The availability of a second valence electron gives Mg (Group 2) the lowest second ionization energy.

If (atomic # is 88) radium-226 were to undergo radioactive decay by electron capture (a type of beta decay) instead of by alpha emission, the resulting nucleus would be: A) 222 Rn 86 B) 226 Fr 87 C) 226 Ra 88 D) 226 Ac 89

B - In all three forms of beta decay, the mass number remains unchanged, while the atomic number increases (β−-decay) or decreases (β+-decay and electron capture) by 1. As the atomic number changes, the identity of the element changes accordingly. If radium-226 underwent an electron capture, the result would yield a nucleus of the same mass number (226) but an atomic number that is 1 less (88 − 1 = 87). Therefore, francium-226 would be detected.

The difference between the first and second ionization energies for magnesium is approximately 700 kJ/mol. The difference between the second and third ionization energies will likely be: A) Higher because the first and second ionization energies remove electrons from the p orbitals whereas the third removes an electron from an s orbital B) Higher because the first and second ionization energies remove valence electrons whereas the third removes a core electron C) Lower because the first and second ionization energies remove spin-paired electrons whereas the third removes an electron that is not spin-paired D) Lower because the first and second ionization energies remove electrons from the n = 2 subshell whereas the third removes an electron from the n = 3 subshell

B - Ionization energy is the energy needed to remove an electron from a ground state gaseous atom or ion. The first, second, and third ionization energies are the energies required to remove the first, second, and third electron, respectively. Because the valence shell is farther away from the positively charged nucleus, electrons in this shell are easier to remove than the inner core electrons. The electrons are removed in order beginning with the electron of highest energy.

What happens to the pOH of the gastric acid entering the duodenum (pH 6.0) as a result of the introduction of bicarbonate ions from incoming pancreatic secretions? A) The pOH increases as [OH−] increases. B) The pOH decreases as [OH−] increases. C) The pOH increases as [OH−] decreases. D) The pOH decreases as [OH−] decreases.

B - Lower pOH values indicate more alkaline solutions with higher [OH−]. This is opposite to the pH scale, in which lower pH values indicate more acidic solutions. Consequently, pH and pOH are inversely correlated. As pH rises, pOH falls (and vice versa) while maintaining the relationship pH + pOH = 14. The bicarbonate ion (HCO3−) is a base (proton acceptor) capable of neutralizing gastric acid through the formation of carbon dioxide and water: H+ + HCO3− → CO2 + H2O The [H+] decreases as bicarbonate consumes the available H+ from the gastric acid, and the relative [OH−] increases. As a result, the duodenum (pH 6.0) is less acidic than the upper stomach (pH = 1.0). Therefore, going from the highly acidic stomach to the less acidic duodenum gives a higher pH and a lower pOH, meaning [H+] decreases and [OH−] increases.

The radioisotope gallium-68 is used in nuclear medicine as a traceable radioactive emitter for diagnostic imaging procedures. The 68Ga isotope is generated from 68Ge by: A) Adding a proton to the nucleus B) Converting a proton in the nucleus to a neutron by electron capture C) Converting a neutron in the nucleus to a proton by β− decay D) Removing three electrons from the atom

B - Nuclear conversions may change the atomic number (number of protons) or the mass number (combined number of protons and neutrons). In beta decay, a proton can convert to a neutron (or vice versa), leaving the mass number unchanged.

At a given temperature, the resistance of a wire to direct current depends only on the: A) Voltage applied across the wire. B) Resistivity, length, and cross-sectional area. C) Inductance, length, and cross-sectional area. D) Resistivity, length, and capacitance

B - R = P L/A R is resistivity

Atoms of a given element have a defined number of protons, neutrons, and electrons, but atoms can gain or lose electrons to form ions. How many neutrons and electrons are present in a 80Se2− ion formed from a neutral selenium-80 atom? A) 80 neutrons and 36 electrons B) 46 neutrons and 36 electrons C) 34 neutrons and 34 electrons D) 46 neutrons and 34 electrons

B - Selenium-80 has a mass number (protons plus neutrons) of 80. Because all atoms of selenium have 34 protons, the number of neutrons is found by subtracting the atomic number (protons) from the mass number (80 − 34 = 46). A neutral (uncharged) atom of selenium-80 in its elemental state has 34 protons and 34 electrons. However, the −2 charge in the 80Se2− ion indicates that the 80Se atom has gained two additional electrons during ion formation. Therefore, a 80Se2− ion has 46 neutrons and 36 electrons. Isotopes are specified by their mass number (the number of protons and neutrons in the nucleus). The number of neutrons in an isotope can be determined by subtracting the element's atomic number (number of protons) from the mass number. An atom in its elemental state is neutral (equal number of protons and electrons) but acquires a net charge of +1 for each electron lost and −1 for each electron gained during ionization.

Why can the positive ions be considered to be fixed during the electrons' oscillations? A) The ions are bound together with strong nuclear forces. B) An ion is much more massive than an electron. C) The ions experience no force when the electron sea is displaced. D) Coulomb's law prohibits the motion of the ions.

B - Since the ions are thousands of times more massive than the electron, answer B is justified (the hydrogen ion is the lightest ion and is nearly 2000 times more massive than the electron). An ion with so much mass compared to an electron will not be able to respond quickly because of its inertia.

The conjugate acid of the dihydrogen orthosilicate anion (H2SiO42−) is: A) HSiO43− B) H3SiO4− C) H4SiO4 D) H3O+

B - The conjugate of an acid (or base) is the species that the acid (or base) becomes after losing (or gaining) one proton. A base is converted into its conjugate acid when the base accepts a proton. Likewise, an acid is converted into its conjugate base when the acid donates a proton. Therefore, an acid (or base) and its conjugate have chemical formulas that differ by only one proton.

If the electromotive force of the battery in an AED is found to be −2.0 V while it is charging, the battery is functioning as a: A) Galvanic cell B) Electrolytic cell C) Concentration cell D) Fuel cell

B - The electromotive force Eocell for a cell is the difference between the standard reduction potential of the reaction at the cathode and the anode. The standard reduction potential of a particular metal or molecule is the potential (in volts) required to reduce the compound. The more negative the value of Eo for any given compound, the less likely it is to be reduced. The equation below is used to determine the standard potential for an oxidation-reduction pair for a particular electrochemical cell: E∘cell=E∘cathode−E∘anode For an electrolytic cell, this value is negative, indicating that the oxidation-reduction reaction is not spontaneous. During the charging phase of a rechargeable battery, an external potential is applied to force the oxidation-reduction reaction to proceed in a nonspontaneous direction. Because Eocell for this battery is a negative value (−2.0 V), the cell is functioning as an electrolytic cell. (Choice A) A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy via an oxidation-reduction reaction. This reaction occurs spontaneously, and Eocell is a positive value. (Choice C) A concentration cell is a type of galvanic cell in which ions diffuse across a membrane, creating a potential. Because concentration is the driving force behind electron transfer, the standard potential of a concentration cell is positive whereas the concentration is different between the two compartments of the cell until it reaches equilibrium. (Choice D) A fuel cell is a type of galvanic cell whose reactants are continuously supplied at the anode and cathode, and whose products are continuously removed from the system. Although outside material is continuously supplied to the fuel cell, its reaction is still spontaneous, meaning that the Eocell is positive. Educational objective:When a battery is charging, it functions as an electrolytic cell because an external potential forces the oxidation-reduction reaction to proceed in a nonspontaneous direction. This makes the electromotive force for the cell a negative value.

Barbiturates are known to inhibit complex I but do not affect complexes II-IV. Based on this fact, how will barbiturates affect the proton motive force? A) The proton motive force increases B) The proton motive force decreases C) The proton motive force is unchanged D) The proton motive force becomes zero

B - The proton motive force (pmf) arises as hydrogen ions (protons) are pumped from the mitochondrial matrix to the intermembrane space by complexes I, III, and IV. It is the driving force behind ATP synthesis in the mitochondria. When complex I is inhibited, the ETC can still operate from complex II onward as long as FADH2 can continue to provide complex II with electrons. Complexes III and IV continue to pump protons, and the pmf remains above zero (Choice D). Because complex I normally participates in proton pumping, its inhibition will result in fewer protons being pumped across the mitochondrial membrane. Therefore, the H+ concentration gradient is reduced and the pmf decreases, resulting in decreased ATP synthesis.

Which of the following properties of β-lactams like those described in the passage best explains their reactivity? A) sp2 hybridization of nitrogen in β-lactam B) Angles less than 109.5° in the β-lactam ring C) Trigonal planar geometry of the nitrogen atom D) Increased resonance

B - β-lactams are a class of four-membered cyclic amides, and a number of factors contribute to their enhanced reactivities relative to noncyclic amides. The nitrogen atom in a noncylic amide is sp2 hybridized because it donates electron density to the carbonyl carbon through resonance. Resonance gives the amide bond partial double-bond character and stability by better distributing electron density and delocalizing charges. The partial double-bond character makes the nitrogen planar. In addition, the small ring size of a β-lactam forces the bond angles to be smaller than the 109.5° normally observed in sp3 hybridized atoms, causing significant ring strain. This strain makes the carbonyl carbon more reactive than it would be in a noncyclic amide.

Lipids are an important component of cell membranes. Which pair of terms lists a class of lipid that is necessary to maintain membrane fluidity and one that is used as markers for cell signaling, respectively? A) Triacylglycerol, cholesterols. B) Phospholipids, waxes. C) Cholesterols, glycosphingolipids. D) Sphingolipids, cholesterols.

C - Cholesterols are cyclic molecules that exist in the phospholipid by layer to maintain fluidity and shape, without them, the bilayer would be too rigid. In contrast, glycosphingolipid are sphingolipids with a sugar attached to the head group. These molecules protrude from the membrane to act as tags for cell signaling. While phospholipids are the predominant component of cell membranes, they contribute to rigidity, not fluidity. Sphingolipids are used to protect cells from the external environment.

Which of the following molecular geometries could describe a compound with a central atom that is dsp3 hybridized? A) Tetrahedral B) Octahedral C) Trigonal bipyramidal D) Square planar

C - Compounds with a dsp3 hybridized central atom have five electron-dense areas, made from one d, one s, and three p orbitals. To minimize repulsion, three of the hybrid orbitals lie in the same plane, 120° apart, and the other two orbitals are 90° above and below the plane. Compounds with dsp3 hybridization and no lone pairs are trigonal bipyramidal. If a compound with dsp3 hybridization has either one or two lone pairs, it will have a see-saw or T-shaped geometry, respectively. (Choice A) Compounds with tetrahedral molecular geometry have four electron-dense areas around an sp3 hybridized central atom. (Choices B and D) Compounds with octahedral (no lone pairs) and square planar (two lone pairs) molecular geometries have six electron-dense areas around a d2sp3 hybridized central atom.

For the electron configuration of iron, Fe(0), do electrons fill the 4s orbitals before the 3d orbitals are filled? A) Yes, because electrons fill the s and p orbitals before they fill the d orbitals, even if they are in different energy levels B) No, because 3d orbitals are lower in energy than 4s orbitals C) Yes, because according to the Aufbau principle, electrons fill lower energy levels first D) No, because orbitals are filled based on the order of the principal quantum number, and 3d comes before 4s

C - Each orbital is associated with a different energy level, and the Aufbau principle states that electrons will fill the lowest energy levels first. The periodic table is arranged to reflect the relative energies of each orbital type and, therefore, the order of electron orbital filling. The orbitals are arranged into different blocks on the periodic table: s orbitals are in the first two columns (except helium), the p orbitals are on the right, the d orbitals are in the middle, and the f orbitals are at the bottom. Therefore, the order of orbital filling can be determined simply by finding an atom's position on the table. Iron is within the 3d orbital section, which comes after the 4s section, so 4s is filled first.

What causes duplex DNA with a certain (A + T):(G + C) ratio to melt at a higher temperature than comparable length duplex DNA with a greater (A + T):(G + C) ratio? A) Stronger van der Waals forces of pyrimidines B) Stronger van der Waals forces of purines C) Increased π- stacking strength D) Reduced electrostatic repulsion of phosphates

C - G-C base pairs form stronger π-stacking interactions than A-T base pairs, thereby creating the most thermal stability. This disparity has often been used to explain the increased melting temperature of DNA rich in GC content

Which separation technique is optimal for purification of small sample sizes and employs a stationary phase, a solvent mobile phase (under pressure), and a detector to separate compound mixtures based on polarity? A) Gas chromatography B) Extraction C) High-performance liquid chromatography D) Thin-layer chromatography

C - High-performance liquid chromatography is a purification technique that separates compounds based on polarity and consists of a mobile phase, stationary phase, detector, and computer for data acquisition. The mobile phase is a solvent or mixture of solvents that are pumped through the system under pressure, and the stationary phase is a column made of either a nonpolar or polar material. Thin-layer chromatography (TLC) separates small sample sizes and includes a stationary phase, solvent mobile phase, and detector (UV). However, TLC is not done under pressure.

The copper(II) acetylsalicylate product immediately precipitates from the reaction solution because: 4 NaC9H7O4(aq) + 2 CuSO4(aq) → Cu2(C9H7O4)4(s) + 2 Na2SO4(aq) A) [Cu2+]2 multiplied by [C9H7O4−]4 is less than the Ksp of Cu2(C9H7O4)4. B) The sum of [Cu2+]2 and [C9H7O4−]4 is less than the Ksp of Cu2(C9H7O4)4. C) [Cu2+]2 multiplied by [C9H7O4−]4 is greater than the Ksp of Cu2(C9H7O4)4. D) The sum of [Cu2+]2 and [C9H7O4−]4 is greater than the Ksp of Cu2(C9H7O4)4.

C - If the concentrations of [X] and [Y] are such that [X]c[Y]d< Ksp, the solution has not reached the solubility limit and will permit more ions into solution (dissolve more of Z). Conversely, if the concentrations of [X] and [Y] are such that [X]c[Y]d >Ksp, the solubility limit of the solution has been exceeded and the solution will not accept any more ions (no more Z will dissolve). This will cause any ions in excess of theKsp to come out of solution as a precipitate.

A glass fiber carries a light digital signal long distances with a minimum loss of amplitude. What optical property of glass allows this phenomenon? A) Dispersion B) Refraction C) Reflection D) Diffraction

C - Light can be carried along a distance within a transparent material by means of total internal reflection.

Which of the following statements best explains why the intensity of sound heard is less when a wall is placed between a source of sound and the listener? A) Sound travels more slowly in a solid than in air. B) The frequency of sound is lower in a solid than in air. C) Part of the sound energy is reflected by the solid. D) The wavelength of sound is shorter in a solid than in air.

C - Sound wave are reflected and absorbed when incident on materials just as light waves are. In doing so the waves are reduced in their forward intensity (units of J/m2). When a wall is placed between a source of sound and the listener, some of the sound energy is transmitted to the listener, but some is reflected and sent backward toward the source or absorbed in the material. The listener hears less sound, but at the same frequency and wavelength as the unimpeded sound.

If the ETC is a spontaneous process, which of the following MUST be true about the redox reactions within each of the four complexes? A) The electron-carrying molecules in the ETC have E0′ values greater than that of oxygen B) The overall free energy (ΔG°′) of the ETC is positive C) The E0′ values increase as electrons are passed from one carrier to the next D) The redox reactions in complex IV are faster than the reactions in complex I

C - Standard reduction potentials are indicative of whether a molecule will spontaneously gain or lose electrons. In a spontaneous process, ΔE0′ is positive. The ETC is a series of spontaneous processes, so all of the standard reduction potentials must fall within the range of NAD+ (E0′ = −0.320 V) and O2 (E0′ = 0.816 V), and each step in the chain must have a larger ΔE0′ than the previous step.

Litmus paper will turn blue when immersed in all of the following EXCEPT: A) Ca(OH)2 B) NH3 C) HNO3 D) NaCN

C - The Brønsted-Lowry definitions of acids and bases can be used to identify acidic and basic molecules. Brønsted-Lowry acids donate a proton (H+ ion), and Brønsted-Lowry bases accept a proton. The negatively charged ions OH− and CN− in Ca(OH)2 and NaCN, respectively, can accept a proton (Choices A and D). The N atom in NH3 (Choice B) has a lone pair of electrons, which can accept an H+ ion. Because these molecules are bases (proton acceptors), they will turn litmus paper blue. HNO3 can donate a proton, making it an acid. Therefore, when litmus paper is immersed in HNO3, it will turn red rather than blue.

In designing the experiment, the researchers used which type of 32P labeled ATP? A) α32P-ATP B) β32P-ATP C) γ32P-ATP D) δ32P-ATP

C - The answer to this question is C because the phosphoryl transfer from kinases comes from the γ-phosphate of ATP. Therefore, the experiment should require γ32P-ATP.

Which circuit elements store energy? I. Capacitors II. Resistors III. Batteries A) I only B) I and II only C) I and III only D) II and III only

C - The capacitor charges up and stores energy in the electric field between the places. The energy stored is ½CVc2, where Vc is the voltage across the capacitor. The battery is the source of energy for the circuit and thus is a store of energy. The resistor is not a storage device for energy and answer C is the correct answer.

In addition to calcium oxalate stones, researchers might hypothesize that potassium citrate (a weak base) could reduce the risk of uric acid stones for which of the following reasons? A) Potassium reduces the hydroxide concentration in urine B) Potassium increases the hydroxide concentration in urine C) Citrate reduces the hydronium concentration in urine D) Citrate increases the hydronium concentration in urine

C - The pH of a solution is defined as the negative logarithm of the hydronium concentration, often expressed in a simplified manner as −log [H+]. pH is related to the negative logarithm of hydroxide concentration (pOH) by the equation pH = 14 − pOH. Therefore, pH can be increased by adding hydroxide ions or by removing hydronium ions. The passage states that citrate alkalinizes (increases the pH of) the urine, and Table 1 indicates that uric acid becomes more soluble with increasing pH. The more soluble a compound is, the less likely it is to precipitate and form kidney stones. Citrate is a weak base and therefore a proton acceptor. It can alkalinize a solution by removing protons from hydronium ions, converting those ions to water and reducing the total hydronium concentration. Therefore, citrate may increase the solubility of uric acid by reducing the hydronium concentration and may decrease the risk of uric acid kidney stones. Increasing the hydroxide concentration requires deprotonation of water to form hydroxide. Potassium is positively charged and would repel positively charged protons, and so be unable to deprotonate water.

A reaction between a long-chain anhydride and phenylethylamine is done to produce a long-chain amide and a carboxylic acid. Which of the following aqueous solutions can separate the products of this reaction in an extraction? A) NaHSO4 B) HNO3 C) LiOH D) HClO4

C - To separate long-chain amides from carboxylic acids in the organic layer, a base must be added. Bases deprotonate the carboxylic acid and generate carboxylate anions that are more soluble in the aqueous layer than in the organic layer. Dilute LiOH is strong enough to deprotonate a carboxylic acid but not to deprotonate an amide N-H. Therefore, addition of the base LiOH will produce carboxylate anions that enter the aqueous layer and allow long-chain amides to remain in the organic layer. (Choice A) NaHSO4 (pKb ∼12) is not basic enough to deprotonate the carboxylic acid. NaHSO4 is the salt of the conjugate base of sulfuric acid (a strong acid), and strong acids produce weak conjugate bases. (Choices B and D) HNO3 and HClO4 are both acids. Acids will not deprotonate carboxylic acids, and therefore cannot be used to separate long-chain amides and carboxylic acids.

Glycoprotein might have an oligosaccharide chain compose of all the following EXCEPT: A) Manose. B) Glucose. C) Deoxyribose. D) Galactose.

C - deoxyribose is a component of DNA. It is not found in the oligosaccharide chains of glycoproteins. The other listed monosaccharides are common components of these chains.

In most forms of gas chromatography: A) The stationary phase is a gas, such as helium, while the mobile phase is a liquid and a solid, inert support structure. B) The stationary phase is a gas, such as hydrogen and oxygen, while the mobile phase is a liquid on a solid, inert support structure. C) The mobile phase is a gas, such as helium, while the stationary phase is a liquid on a solid, inert support structure. D) The mobile phase is a gas, such as hydrogen or oxygen, while the stationary phase is a liquid on a solid, inert support structure.

C - most gas chromatography conducted in organic chemistry lab is gas liquid chromatography. The liquid comprised of the stationary phase, while an inner gas like helium or nitrogen passage to the apparatus of the mobile phase.

Which of the following describes the direction of the magnetic force on an ion moving in an artery past a flowmeter? A) Parallel to both the direction of v and the direction of B B) Parallel to the direction of v and perpendicular to the direction of B C) Perpendicular to the direction of v and parallel to the direction of B D) Perpendicular to both the direction of v and the direction of B

D - A magnetic force acts on a moving charge in a direction that is perpendicular to both the velocity of the charge and the direction of the magnetic field. This is a basic law of the interaction of electric currents and magnetic fields.

According to trends in ionization energy, which metal would be the most reactive with Group 6A and 7A nonmetals? A) Li B) Na C) K D) Cs

D - Alkali metals, or Group 1 metals, are known to be very reactive because of their low first ionization energy, large atomic radius, and smallelectronegativity. The alkali metals have one electron in their valence shell, which they readily donate to form strong ionic bonds with nonmetals such as the halogens (Group 7A, or Group 17). As the atoms in the alkali metal group increase in size, the valence electrons occupy higher energy levels, and less energy is required to remove an electron from the valence shell. Therefore, larger alkali metals are more reactive. Cesium (Cs) has the lowest first ionization energy of all the elements. Consequently, less energy is required to remove an electron from its valence shell compared to other metals. As a result, Cs is the most reactive metal in reactions with nonmetals.

Which of the following statements describe a characteristic of an ideal gas? I. There are no attractive or repulsive forces between the gas molecules. II. The size (molecular volume) of the individual gas molecules is negligible. III. Collisions between the gas molecules are completely elastic. A) I and II only B) I and III only C) II and III only D) I, II, and III

D - An ideal gas has four characteristics: An ideal gas has no attractive or repulsive forces between the gas molecules (Number I). The size (molecular volume) of the individual gas molecules of an ideal gas is negligible (taken to be zero) compared to the volume (space) of the container the gas occupies (Number II). Collisions between the molecules of an ideal gas are completely elastic (no energy is lost by interactions or friction) (Number III). Ideal gas molecules have an average kinetic energy (energy of motion) that is directly proportional to the gas temperature.

Two separate reactions are conducted in which a compound containing a ketone, an ester, and a carboxylic acid is reacted with borane (BH3) in THF in one reaction and with NaBH4 in methanol in the other. Which of the following explains why different products are observed? A) BH3 will selectively reduce ketones, and NaBH4 will only reduce carboxylic acids. B) BH3 does not reduce ketones, and NaBH4 will selectively reduce esters. C) BH3 reduces carboxylic acids and esters, and NaBH4 will only reduce esters. D) BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce ketones.

D - BH3 reacts most readily with carboxylic acid carbonyl groups and therefore will selectively reduce carboxylic acids. As a result, the intermediate aldehyde will be further reduced to a primary alcohol. NaBH4 reduces the reactive carbonyls—ketones and aldehydes

Consider the trends in atomic radii on the periodic table. The bond in citrate that is the longest and weakest with the lowest bond dissociation energy is the: A) C-H bond B) C=O bond C) O-H bond D) C-C bond

D - Bond dissociation energy is the energy required to break a chemical bond; it is related to the bond length, or the distance between the nuclei of two bonded atoms. Shorter bonds are stronger and require more energy to break. Atoms with small atomic radii can form short, strong bonds whereas atoms with larger radii form longer, weaker bonds. Each subsequent row on the periodic table indicates an additional electron shell, and therefore a larger atomic radius, so atoms in the first row will have smaller atomic radii than atoms in the second row, and so on. Double bonds require more energy to break than single bonds. Citrate includes bonds between carbon and carbon, carbon and oxygen, carbon and hydrogen, and oxygen and hydrogen. Of the choices given, the C-C bond is the only single bond that does not involve a first row atom (hydrogen). Therefore, the C-C bond will be the longest and require the least energy to break.

The electrolysis of aqueous silver nitrate involves two separate half-reactions that occur at the anode (Reaction 1) and the cathode (Reaction 2). 2H2O(l) → O2(g) + 4H+(aq) + 4e− Reaction 1 Ag+(aq) + e− → Ag(s) Reaction 2 Which of the following net reactions occurs during the electrolysis? A) 2H2O(l) + Ag+(aq) → O2(g) + Ag(s) + 4H+(aq) B) 2H2O(l) + 4Ag+(aq) → O2(g) + 4Ag(s) + 2H2(g) C) 2H2O(l) + Ag(s) → O2(g) + Ag+(aq) + 4H+(aq) D) 2H2O(l) + 4Ag+(aq) → O2(g) + 4Ag(s) + 4H+(aq)

D - Canceling the 4e− terms and summing the reactants side and the products side of the half-reactions yield a net reaction with both an atom balance and a net charge balance (+4 on each side).

Making which of the following changes to a circuit element will increase the capacitance of the capacitor described in the passage? A) Replacing the 500-Ω resistor with a 250-Ωresistor B) Replacing the 10-V battery with a 20-V battery C) Increasing the separation of the capacitor plates D) Increasing the area of the capacitor plates

D - Capacitance C depends on geometric factors only, and in the case of parallel plates, C is proportional to the plate area and inversely proportional to the separation distance of the plates. Thus, D is the best answer.

Conversion of a keto group to a ketone could be accomplished by adding: A) Potassium tert-butoxide. B) Lithium aluminum hydride. C) Sodium hydroxide. D) Aqueous sulfuric acid.

D - Ketals may be effectively hydrolyzed by treatment with strong acid and water. B: While lithium aluminum hydride is a powerful reducing agent it is unlikely to have an effect on the ketal. A,C: Ketals and acetals are resistant to basic conditions; while both are acid sensitive.

When a light wave and a sound wave pass from air to glass, what changes occur in their speeds? A) Both speed up. B) Both slow down. C) Light speeds up; sound slows down. D) Light slows down; sound speeds up.

D - Light slows down because the index of refraction in the glass is greater than in the air. The index is a measure of the ratio of the velocity in air to the velocity in the medium. For sound the speed becomes greater because the speed of sound in a solid is much greater than in air (the glass has stiff rigid bonds which gives rise to a speed more than 10 times greater than in air).

Which additional extraction steps would cause phosphatidylethanolamine and 2,6-dimethoxyphenol to enter the aqueous layer consecutively? A) Add 0.05 M NaOH(aq) to the organic layer followed by 0.01 M H2SO4(aq). B) Add 0.01 M NaHCO3(aq) to the organic layer followed by 0.05 M HCl(aq). C) Add 0.01 M H2CO3(aq) to the organic layer followed by 0.05 M NaHCO3(aq). D) Add 0.05 M H2SO4(aq) to the organic layer followed by 0.01 M NaOH(aq).

D - Strong organic bases form ionic salts easily when weak acids are added to a mixture. On the other hand, weaker bases such as amines are only protonated by strong acids. Likewise, carboxylic acids, which are considered relatively strong organic acids, can be deprotonated by strong and weak bases. However, phenols are much weaker acids that require strong bases to be deprotonated. For phosphatidylethanolamine and 2,6-dimethoxyphenol to enter the aqueous layer consecutively, their ionic salts must be formed in sequence. Phosphatidylethanolamine contains an amine group that can be protonated to form a water-soluble ammonium salt; therefore, a strong acid such as H2SO4 or HCl should be added to protonate the amino group. In contrast, 2,6-dimethoxyphenol has a weakly acidic hydroxyl group on the benzene ring that must be deprotonated to form an ionic salt. Only strong bases such as NaOH or KOH are able to remove protons from phenols. Consequently, the correct sequence of extraction steps would involve the addition of a strong acid followed by a strong base.

When a weak acid (HA) is titrated with sodium hydroxide in the presence of an indicator (HIn), the pH at which a color change is observed depends on the: A) Final concentration of HA. B) Final concentration of HIn. C) Initial concentration of HA. D) pKa of HIn

D - The indicator will change color over a specific pH range. The range at which the color change takes place depends on the point at which HIn is converted to In-, and this depends on the pKa of the indicator which is answer D.

According to the passage, the energy released when an atom splits comes from: A) Fast-moving electrons. B) The short-range attraction of the nucleons. C) Mutual attraction of the fragments. D) Mutual repulsion of the fragments.

D - The kinetic energy of the fission particles is produced by the repulsive Coulomb force between the particles. Thus, answer D is the best choice.

Increasing the frequency of each photon that is directed at the cathode will: A) Decrease the number of photons ejected. B) Increase the number of photons ejected. C) Decrease the speed of the ejected electrons. D) Increase the speed of the ejected electrons.

D - The only effect the photon frequency has on the ejected electron is on its kinetic energy. Photon energy equals cathode work function plus electron kinetic energy. The number of electrons ejected (the current) depends on the number of incident photons.

Given that acetic acid (CH3COOH) has pKa ≈ 5, the estimated pH of an aqueous 0.001 M sodium acetate (CH3COONa) salt solution is: A) < 4 B) > 4 and < 7 C) = 7 D) > 7

D - The tendency of the salt ions to act as an acid or a base depends on the strengths of the acid and the base involved in forming the salt. Strong acids form weak conjugate bases, and strong bases form weak conjugate acids. Therefore, examining the strengths of the ionic species permits a qualitative assessment of the salt solution pH: Neutral salts formed from a strong acid reacting with a strong base give neutral solutions (pH = 7). Acidic salts formed from a strong acid reacting with a weak base give acidic solutions (pH < 7). Basic salts formed from a weak acid reacting with a strong base give basic solutions (pH > 7).

And unsaturated free fatty acid must contain: I. An alkene or alkyne. II. An esther. III. A carboxylic acid. A) I only B) III only C) I and II D) I and III

D - The term unsaturated indicates that there is at least one multiple bomb, generally an alkene, in the molecule. Additionally, any fatty acid, as it's name implies, must have a carboxylic acid at one terminal.

Assume that a NiCd battery charges for 30 minutes (t) with an applied current (I) of 5.0 A. Which expression could be used to calculate the number of moles of Cd produced on the cathode from electrolytic plating? A) nF/It B) FI/nt C) I/tnF D) It/nF

D - To determine the number of moles of Cd plated on the cathode, first determine the charge Q (coulombs, C) that flowed through the NiCd battery. The charge is the product of the current I (amperes, or C/s) and time (t, seconds), given by the expression: Q=It Next use the Faraday constant F to relate the charge of the electrolytic cell to moles of electrons n needed to reduce the metal. The charge per mole of Cd is the product of F and the moles of electrons n needed to reduce cadmium, resulting in the equation: It/mol Cd=nF Solving for moles of Cd results in the equation: mol Cd=It/nF

Fisher Esterification

Esters are carboxylic acid derivatives that can be formed under acidic conditions through a Fischer esterification using a carboxylic acid and an alcohol. The reaction begins with protonation of the carbonyl oxygen, followed by nucleophilic attack of the carbonyl carbon by the alcohol. A proton is then transferred from the alcohol oxygen to the hydroxyl group of the carboxylic acid, resulting in water (a good leaving group). Lastly, the carbonyl oxygen is deprotonated to yield an ester.

Time in flight

Time in flight is determined by vertical, not horizontal, components. Increasing the total velocity while leaving the angle unchanged must also increase the vertical velocity, and this time.

What is the advantage of vacuum distillation over simple distillation?

Vacuum distillation facilitates separation of compounds with extremely high boiling points. A compound will boil when its vapor pressure is equal to the ambient pressure to which it is exposed. Adding a vacuum to the distillation apparatus dramatically lowers ambient pressure, make a compound that typically have high boiling points more prone to vaporization. For example, consider a compound that boils at 300°C at one ATM of pressure, this substance will be nearly impossible to vaporize without using a vacuum.

Electrolytic Cell

When a battery is charging, it is an electrolytic cell, and when it is discharging, it is a galvanic cell. The difference between charging and discharging is determined by which electrode is the cathode and which is the anode. Although electron flow is in the opposite direction when a battery is charging as when it is discharging, electrons always flow from the anode to the cathode. Reduction occurs at the cathode, and oxidation occurs at the anode.


संबंधित स्टडी सेट्स

Cumulative Exam: Physical Assessment

View Set

Section 8: The Sales Contract in Texas

View Set

Stanhope Ch 1, Stanhope Ch. 12: Communicable and Infection Disease Risks, Chapter 14: Communicable and Infectious Disease Risks; Stanhope, Chapter 14: Communicable and Infectious Disease Risks (Stanhope: Public Health Nursing, 8th Ed), Chapter 12: Ep…

View Set